Let denote the number of subsets of . Consider the power set of . Partition set into two blocks and . Clearly, we have that . And, with the exception of we may (and can) only put into each of the subsets of in . Thus, we have that ...almost...we double counted one thing (what is it??). So . Thus, it follows by induction that the recurrence relation has solution for some . Noting though that is the number of subsets of the empty set we may conclude that . The conclusion follows.

Why so complicated? Just notice that if you want to construct a subset of , you have two possibilities for every element of : either you include it, or you exclude it. So in total you have possibilities.

Why so complicated? Just notice that if you want to construct a subset of , you have two possibilities for every element of : either you include it, or you exclude it. So in total you have possibilities.

Your proof relies on induction whereas my proof relies on a combinatorial argument... I don't think they're similar at all!

The induction is merely the formality of it. The point was that if you fix an element in it only has two places to go. I really do think we are saying the exact same thing, just the means of conveyance is different.

Why so complicated? Just notice that if you want to construct a subset of , you have two possibilities for every element of : either you include it, or you exclude it. So in total you have possibilities.

Another proof : clearly the number of subsets of is

The "another proof" is not so unless you first prove that there are subsets with k elements out of a set with n elements, , and also you forgot the number in the sum and also, and perhaps most important, this another proof relies on "hidden induction": after all, we need to show that's true for all n, just like the formal proof of Newton's binomial theorem is usually, and as far as I am aware uniquely, done by induction.